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- anonymous

Suppose ΔABC is an acute triangle. Let the bases of the altitudes at B and C be E and F, respectively. Prove ΔABC and ΔAEF are similar

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- anonymous

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- anonymous

I typed the question verbatim, but it sounds odd that E and F would be bases. I would think they would be points. Regardless of that, still am unsure of how to do the problem.

- ganeshie8

|dw:1437110964213:dw|

- anonymous

Oh, okay, so they really are supposed to be points E and F. Maybe I drew my picture wrong but, at least the way yours is drawn, you could maybe show FE is parallel to BC?

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- ganeshie8

That is equivalent, but there is a simpler way here

- anonymous

Oh, thats equivalent to saying theyre similar? Saying those lines are parallel?

- ganeshie8

First establish trivially that \(\triangle OFB\sim\triangle OEC\) by \(AA\)
|dw:1437112233723:dw|

- anonymous

Right, made it that far at least xD

- ganeshie8

Next show that \(\triangle AFC\sim AEB\) by AA
|dw:1437112450857:dw|
actually we don't need the first step :)

- ganeshie8

setup a proportion and we're done!

- anonymous

Alright, I have that similarity. After that, maybe because I was being confused by my picture, I wasnt sure. And we're setting up a proportion based on the sides of triangle AFC and triangle AEB?

- ganeshie8

Yes :
\[\dfrac{AF}{AE}=\dfrac{AC}{AB}\]
together with \(\angle A\cong \angle A\) by reflexive property yield \(\triangle ABC\sim\triangle AEF\) by \(SAS\) similarity

- anonymous

Well, we didn't really talk about a lot of the high school geometry postulates and such. It seems like the professor just assumes that what we are given implies everything else. Maybe what we were given implies SAS similarity? This is all we were given about similar triangles:
"ΔABC ~ ΔA'B'C' iff \(\frac{|A'B'|}{|AB|} = \frac{|A'C'|}{|AC|} = \frac{|B'C'|}{|BC|}\)"
Or
"ΔABC ~ ΔA'B'C' iff the angle at A = angle at A', angle at B = angle at B', and angle at C = angle at C'"

- ganeshie8

"ΔABC ~ ΔA'B'C' iff the angle at A = angle at A', angle at B = angle at B', and angle at C = angle at C'"
This is equivalent to AA similarity because the third angle is determined by triangle sum property once the first two angles are known.

- ganeshie8

"ΔABC ~ ΔA'B'C' iff \(\frac{|A'B'|}{|AB|} = \frac{|A'C'|}{|AC|} = \frac{|B'C'|}{|BC|}\)"
This is SSS similarity, lets see if we can jump to this from SAS

- ganeshie8

here it is, pls scroll directly to SAS similarity
http://ceemrr.com/Geometry1/SSSandSASsimilarity/SSSandSASsimilarity_print.html
they're proving SAS similarity using AA

- anonymous

Alrighty, Ill take a look at that and make sure I see the SAS. Thanks again :)

- ganeshie8

This is analogous to SAS congruence :
two triangles are congruent if the corresponding pairs of two adjacent sides and the included angle are congruent

- ganeshie8

SAS similarity says :
two triangles are "similar" if the corresponding pairs of two adjacent sides are "proportional" and the included angles are congruent.

- anonymous

In the link, the statement they give for SAS says
"If \(\frac{AB}{DE} = \frac{AC}{DF}\) and \(∠A\) congruent to \(∠B\)..."
Yet in their diagram, it looks as if it would need to say that \(∠A\) congruent to \(∠D\)

- ganeshie8

Ahh, I overlooked that. thats definitely a typo yeah

- ganeshie8

|dw:1437114236422:dw|

- anonymous

Okay, cool. I gotta learn to use the latex and drawing tools and such like you know how to, haha.

- ganeshie8

I'm using a small chrome extension that gives plotting/graphing like features to the drawing tool

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